Assume that under specific conditions, hydrogen atoms are permitted...

The wavelength of the emitted photon is given by the Rydberg formula:

\dfrac{1}{\lambda} = R \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

For the smallest wavelength, the energy difference between the transition levels must be maximum. Given the condition that the principal quantum number changes by $2$, we have $n_2 - n_1 = 2$. The maximum energy difference occurs for the lowest possible values of $n_1$ and $n_2$.

Thus, the transition is from $n_2 = 3$ to $n_1 = 1$.

Substituting these values into the formula:

\dfrac{1}{\lambda} =R \left(\dfrac{1}{1^2} - \dfrac{1}{3^2} \right) =R \left( 1 -\dfrac{1}{9} \right) = \dfrac{8R}{9}

$\lambda =$\dfrac{9}{8R}

Using the standard value \dfrac{1}{R} \approx$91.2 \text{ nm}$:

$\lambda =$\dfrac{9}{8} \times$91.2 \text{ nm} \approx 102.6 \text{ nm}$

The closest value among the given options is $103 \text{ nm}$.